Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 32


See the explanation below.

Work Step by Step

$\sin 2t-\cot t=-\cot t\cos 2t$ Recall the Trigonometric Identities and apply below, $\begin{align} & \cot t=\frac{\cos t}{\sin t} \\ & \cos 2t=1-2{{\sin }^{2}}t \\ & \sin 2t=2\sin t\cos t \\ \end{align}$ Consider the right side of the given expression and apply the above formula, $\begin{align} & -\cot t\cos 2t=-\frac{\cos t}{\sin t}\left( 1-2{{\sin }^{2}}t \right) \\ & =-\frac{\cos t}{\sin t}+\frac{\cos t2{{\sin }^{2}}t}{\sin t} \\ & =-\frac{\sin t}{\cos t}+2\sin t\cos t \\ & =-\cot t+\sin 2t \end{align}$ Hence, it is proved that the given identity $\sin 2t-\cot t=-\cot t\cos 2t$ holds true.
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